Question #71e84

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Answer 1 · 2017-07-27T14:27:05

#V_(NH_3) approx 20. L#

This is a stoichiometry-gas law problem.

#N_2+3H_2 to 2NH_3#

#(10L)/(22.4L) = 0.446mol[N_2]#
#5g* (2.02g)/H_2 = 2.48mol[H_2]#

Based on stoichiometry, you can see that #N_2# will be the limiting reactant, so let's calculate the amount at STP.

#1.0atm*10.0L = n*(0.08206L*atm)/(mol*K)*273.15K = n = 0.446#

Well, that was a waste of time, but it's nice to double check that they are the same, anyways:

#0.446mol*(2NH_3)/(N_2) approx 0.892mol[NH_3] = n#

#1.0*V = 0.892*(0.08206L*atm)/(mol*K)*273.15K#
#thereforeV approx 20. L#